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Applications of the
Derivative
Chapter 14
Ch. 14 Applications of the Derivative


14.1 Absolute Extrema
14.3 Business Applications of Extrema
14.1 Absolute Extrema
ABSOLUTE MAXIMUM OR MINIMUM
 Let f be a function defined on some interval. Let c be a
number in the interval. Then f(c) is the absolute
maximum of f on the interval if
f(x)  f(c)
for every x in the interval, and f(c) is the absolute
minimum of f on the interval if
f(x)  f(c)

for every x in the interval.
A function has an absolute extremum (plural: extrema)
at c if it has either an absolute maximum or absolute
minimum there.
Absolute Extrema



Consider the graph of function f on the interval [-4, 4]
Absolute maximum at x = 4 (end point)
Absolute minimum at x = 1
Absolute Extrema
EXTREME VALUE THEOREM
 A function f that is continuous on a closed interval [a, b]
will have both an absolute maximum and an absolute
minimum on the interval. Each of these occurs either at
an endpoint of the interval or at a critical number of f.
Absolute Extrema
FINDING ABSOLUTE EXTREMA
• To find absolute extrema for a function f continuous
on a closed interval [a, b]
1. Find all critical numbers for f in (a, b)
2. Evaluate f for all critical numbers in (a, b)
3. Evaluate f for the endpoints a and b of the interval [a, b]
4. The largest value found in step 2 or 3 is the absolute
maximum for f on [a, b], and the smallest value found
is the absolute minimum for f on [a, b]
Absolute Extrema – example (closed interval)
Find the absolute extrema of the function
f(x) = 2x3 + 3x2 – 12x – 7
on the interval [-3, 0]
Solution:
1. Find critical numbers
f ‘(x) = 6x2 + 6x – 12
= 6(x + 2)(x – 1)
Set f ‘(x) = 0 and solve for x
Critical numbers are x = -2 and x = 1
Only -2 lies in the interval [-3, 0]
2, 3. Evaluate f for x = -2, and for endpoints x = -3 and x = 0.
f(-2) = 13
f(-3) = 2
f(0) = -7
4. Absolute maximum occurs at (-2, 13),
Absolute minimum occurs at (0, -7)
Absolute Extrema – example (closed interval)
Absolute maximum
(-2, 13)
15
10
(-3, 2)
f(x) = 2x3 + 3x2 – 12x – 7
[-3, 0]
5
0
-4
-3
-2
-1
-5
Absolute minimum (0, -7)
-10
-15
-20
0
1
2
3
Application



For several weeks, the highway department has been
recording the speed of freeway traffic flowing past a
certain downtown exit. The data suggest that between
1:00 and 6:00 P.M. on a normal weekday, the speed of
the traffic at the exit is approximately
S(t) = t3 – 10.5t2 + 30t + 20 mph
where t is the number of hours past noon. At what time
between 1:00 and 6:00 P.M. is the traffic moving the
fastest, and at what time is it moving the slowest?
In other words, find the absolute maximum and absolute
minimum of the function S(t) on the interval [1, 6].
Application, solution
S(t) = t3 – 10.5t2 + 30t + 20 mph, [1, 6]
1. Find the critical numbers
S’(t) = 3t2 – 21t + 30
= 3(t2 – 7t + 10)
= 3(t – 2)(t – 5)
Set S’(t) = 0
t = 2, t = 5
2. Evaluate S for t = 2 and t = 5, and for endpoints t = 1 and
t = 6.
S(1) = 40.5 S(2) = 46 S(5) = 32.5 S(6) = 38
Absolute maximum = 46. Traffic is moving fastest at 2:00
Absolute minimum = 32.5. Traffic is moving slowest at
5:00
Application, solution
S(t) = t3 – 10.5t2 + 30t + 20 mph
[1, 6]
Traffic Speed
70
Maximum speed
(2, 46)
60
S(t)
50
40
30
Minimum speed
(5, 32.5)
20
10
0
0
1
2
3
4
t
5
6
7
8
14.3 Business Applications of Extrema
1.
2.
3.
4.
Economic lot size
Economic order quantity
Elasticity of Demand
Profit Maximization
1. Economic Lot Size
 Economic lot size – The quantity of output that allows a
manufacturer to meet the demand for their product
while minimizing total cost.
 Total costs:

Set-up costs

Manufacturing costs

Storage costs
Economic Lot Size


Gadget, Inc. produces 12,000 widgets per year (annual
demand = 12,000). The widgets can be produced in
several batches of equal size per year.
 Small number of large batches
 Reduces set-up costs.
 Increases storage costs.
 Large number of small batches
 Increases set-up costs
 Reduces storage costs
Economic lot size – The number of widgets that should
be manufactured in each batch to minimize total cost.
•Lowest set-up costs
•Highest storage costs
•Highest set-up costs
•Lowest storage costs
Economic Lot Size




Let:
 q = number of units in each batch;
 k = cost of storing one unit of the product for one year;
 f = fixed setup cost to manufacturer the product;
 g= cost of manufacturing a single unit of the product;
 M = total number of units produced annually
Manufacturing cost per batch = f + gq
The number of batches per year = M/q
M
Total annual manufacturing cost =  f  gq 
q
fM  gqM fM


 gM
q
q
Economic Lot Size

Let:
 q = number of units in each batch;
 k = cost of storing one unit of the product for one year;
 f = fixed setup cost to manufacturer the product;
 g= cost of manufacturing a single unit of the product;
 M = total number of units produced annually
Cost of storing one unit for a year = k.
Average inventory = q/2
 q  kq
Total storage cost = k   
2 2
Economic Lot Size

Total production cost = manufacturing cost + storage cost
fM
kq
T q 
 gM 
q
2

To find the value of q that minimizes T(q), find T’(q) and
set T’(q) = 0 (Remember: f, g, k, and m are constants)
 fM k
T 'q  2 
q
2
Set T’(q) = 0
solve for q
 fM k
 0
2
q
2
Economic Lot Size

To find the value of q that minimizes T(q), find T’(q) and
set T’(q) = 0 (Remember: f, g, k, and m are constants)
 fM k
 0
2
q
2
k fM
 2
2 q
2 k
q  fM
2
2 fM
2
q 
k
2 fM
q
k
Economic Lot Size
INVENTORY PROBLEM
Economic lot size that minimizes total production cost
2 fM
q
k
Where q = quantity, f = fixed setup cost, M = total units
produced annually, k = cost of storing one unit
for one year.
Economic Lot Size - example

1.
2.
Suppose 100,000 lamps are to be manufactured
annually. It costs $1 to store a lamp for 1 year, and it
costs $500 to set up the factory to produce a batch of
lamps.
Find the number of lamps to produce in each batch.
Find the number of batches of lamps that should be
manufactured annually.
Solution
1. k = 1, M = 100,000, f = 500
2  500 100, 000 
2 fM
q

k
1
 100, 000, 000  10, 000
M 100, 000
2.

 10
q
10.000
2. Economic Order Quantity
Deals with the problem of reordering an item that is
used at a constant rate throughout the year.
 How often to order?
 How many units to order each time an order is
placed?
q = number of units to order each time
k = cost of storing one unit for one year
f = fixed cost to place an order
M = total units needed per year
 Goal: Minimize the total cost of ordering over a year’s
time, where
Total cost = Storage cost + Reorder cost

Economic Order Quantity



As defined previously, average inventory = q/2, so
yearly storage cost is kq/2
Number of orders placed annually = m/q, so reorder cost
is fM/q
fM kq
Total reorder cost is T  q  

 fM k

and, T '  q  
2
q
2
q
2
When T’(q) is set = 0 and, we solve for q
2 fM
q
k
Same as the “inventory problem”
Example


A bookstore has an annual demand for 100,000 copies of
a best-selling book. It costs $.50 to store 1 copy for 1 year,
and it costs $60 to place an order. Find the optimum
number of copies per order.
Solution
k = 0.5, M = 100,000, f = 60
2 fM
2(100,000)(60)
q

 4,898.98
k
0.5
60 100, 000  0.5  4898 
T  4898 

 2449.489792
4898
2
60 100, 000  0.5  4899 
T  4899  

 2449.489743
4899
2
3. Elasticity of Demand

The responsiveness of the quantity demanded of a good
to a change in its price, ceterus paribus.
ELASTICITY OF DEMAND
Let q = f(p), where q is quantity demanded at a price p.
The elasticity of demand is
p dq
E 
q dp
Demand is inelastic if E < 1
Demand is elastic if E > 1
Demand has unit elasticity if E = 1
Example

1.
Research has indicated that the demand for heroin is
given by q = 100p-0.17
Find E
dq
 17 p 1.17
dp
p dq
p
1.17
E 


17
p

0.17 
q dp 100 p
17 p 0.17
E
 0.17
0.17
100 p
2.
Is the demand for heroin elastic or inelastic?
 0.17 < 1, the demand is inelastic.
Elasticity of Demand
Revenue and Elasticity
•If demand is inelastic, total revenue increases as
price increases.
•If demand is elastic, total revenue decreases as price
increases.
•Total revenue is maximized at the price where
demand has unit elasticity.

For many products, elasticity varies at different price
levels.
Example

a.
2
q

216

2
p
Assume the demand for a product is
,
where p is the price in dollars.
Find the price intervals where demand is elastic and
where demand is inelastic.
Example

a.
2
q

216

2
p
Assume the demand for a product is
,
where p is the price in dollars.
Find the price intervals where demand is elastic and
where demand is inelastic.
2
 Solution Since q  216  2 p , dq dp  4 p , and
p dq
E 
q dp
p

4 p 
2 
216  2 p
4 p2

216  2 p 2
Example

a.
2
q

216

2
p
Assume the demand for a product is
,
where p is the price in dollars.
Find the price intervals where demand is elastic and
where demand is inelastic.
4 p2
E
216  2 p 2
To decide where E < 1 or E > 1, solve the equation E = 1
4 p2
1
2
216  2 p
4 p 2  216  2 p 2
6 p 2  216
p 2  36
At p = $6, demand is
unit elastic.
p6
Example

a.
2
q

216

2
p
Assume the demand for a product is
,
where p is the price in dollars.
Find the price intervals where demand is elastic and
where demand is inelastic.
4 p2
E
216  2 p 2

Substitute a test number on either side of 6 in the
expression for E to see which values make E < 1 and
E>1
4  5
2
216  2  5 
2
 .60  1
47
2
216  2  7 
2
 1.66  1
The demand is inelastic at prices less than $6, and elastic
at prices greater than $6
Profit Maximization
Marginal Analysis Criterion For Maximum Profit
 Profit P(q) = R(q) – C(q) is maximized at a level of
production q where marginal revenue equals marginal
cost; that is, where
R’(q) = C’(q)
Remember the MR=MC Rule?
200
Profit
Per Unit
Price, costs, and revenue
175
150
MC
125
Profit
100
ATC
D
75
50
MR = MC
25
0
1
2
3
4
MR
5
6
7
8
9
10
Q
Profit Maximization - Example





A manufacturer determines that when q thousand
widgets are produced (and sold each month), the total
revenue will be
R(q) = -1.2q2 + 22.2q
and the total cost will be
C(q) = 0.4q2 + 3q + 40
What level of production will maximize profit? Or, at
what level of production does MR = MC?
MR = R’(q) = -2.4q + 22.2
MC = C’(q) = 0.8q + 3
-2.4q + 22.2 = 0.8q + 3
3.2q = 19.2
q=6
Profit Maximization - Example

Perform the first derivative test:
 P(q) = MR – MC
= (-1.2q2 + 22.2q) – (0.4q2 + 3q – 40)
= -1.6q2 + 19.2q – 40
P’(q) = -3.2q + 19.2
P’(6) = -3.2(6) + 19.2 = 0
P’(5) = 3.2 (increase q to increase P)
P’(7) = -3.2 (decrease q to increase P)
P(q) = -1.6q2 + 19.2q – 40
Max profit at q = 6
30
20
10
P(q)
0
-10
0
2
4
6
-20
-30
-40
-50
q
8
10
12
Chapter 14
End